New answers tagged

2

Since $T - \lambda I$ is also normal, we have $$ \| T - \lambda I \| = \text{spr} (T - \lambda I) = 0, $$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)


1

using spectral theorem since $T$ is normal it exist a spectral measure $E$ such that $$ Tx=\int_{\sigma(T)}tdE(x)=\int_{\{\lambda\}}tdE(x)=\lambda E(\{\lambda\})(x)=\lambda E(\sigma(T))(x)=\lambda I (x)=\lambda x $$


-2

completeness in hilbert space means that the metric associated with inner product converges,let draw a 1/x as a function on a graph,you will see its area converges to zero. as x approaches to infinity 1/x goes to zero


3

Yes. For a positive linear operator $A$, $\|A\| = \sup \{(Ax, x): \; \|x\| = 1\}$. If $A \ge B$, $(Ax, x) \ge (Bx, x)$, so we must have $\|A\| \ge \|B\|$.


0

For any subset: $$A\subseteq\mathcal{H}:\quad A\cap A^\perp=\{0\}$$ As well as it is: $$A\subseteq\mathcal{H}:\quad\overline{\langle A\rangle}^\perp=A^\perp=\overline{\langle A^\perp\rangle}$$ Moreover it holds: $$A\subseteq\mathcal{H}:\quad A^{\perp\perp}=\overline{\langle A\rangle}$$


-1

1) is just Pythagoras theorem applied to $f(x)$ 2) is just that you need completness because somewhere you use a cauchy sequence you need to converge 3)What's your definition of $Tr(f)$ ?


2

1) You know that $$f(x)=\sum_{i=1}^\infty \lambda_ie_i$$ where $\lambda_i\in\mathbb R$. Now, since $\{e_i\}$ is an orthonormal basis, you'll get, $$\left< f(x),e_i\right>=\sum_{i=1}^\infty \left<\lambda_j e_j,e_i\right>=\sum_{j=1}^\infty \lambda_j\underbrace{\left<e_j,e_i\right>}_{=\delta_{ij}}=\lambda_i.$$ Notice that $\left<f(x),e_i\...


0

I'm not following you here. So let $H$ be a Hilbert space and $V$ a (not necessarily closed) subspace, then $V^{\perp}=\left\{w\in H\mid \left\langle v,w\right\rangle=0 \mbox{ for all }v\in V\right\}$ is the orthogonal complement. Using the continuity of the inner-product you can show that $V^{\perp}$ is a closed subspace of $H$. It follows that $V^{\perp\...


2

Start with $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^\perp. $$ As you noted, $\mathcal{N}(T^{1/2})\subseteq\mathcal{N}(T)$. Because $T^{1/2}$ is selfadjoint, $$ \|T^{1/2}x\|^2=(T^{1/2}x,T^{1/2}x)=(Tx,x). $$ Therefore $\mathcal{N}(T)\subseteq\mathcal{N}(T^{1/2})$. So, $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^{\perp}=\...


2

Of course as Hilbert spaces $L^2(S^1)$ and $L^2\bigl([0,1]\bigr)$ are isomorphic, and you could also say that $L^2\bigl([0,1]\bigr)$ is the prime example of a Hilbert space arising from Lebesgue theory. But note that $L^2(S^1)$ is one of the most important Hilbert spaces in the world, and there definitively is an essential difference between $L^2\bigl([0,1]\...


3

$L^2$ spaces should not be sensitive to the topology or shape of whatever underlying space you're working with. Indeed, given a "manifold" (a generalization of circles and surfaces), one way of defining an $L^2$ space on it is to pick a chart $D^n \to M$, where $D$ is the unit disc, that is injective except at a set of measure zero, and then pull back ...


1

The "basis" you imply in 4. and 5. is the set of all shifted delta distributions $\{\delta_x (f) = f(x)\,\forall\,f\in \mathscr{S}(\mathbb{R}^N)\}$ where $\mathscr{S}(\mathbb{R}^N)\subset L^2(\mathbb{R}^N)$ is the Schwartz Space of bounded smooth functions $f:\mathbb{R}^N\to\mathbb{R}^N$ such that any derivative $D^\alpha (p\,f)$ of any multiple $f p$ of $f$ ...


3

The `basis' you described in 3 is given by the Dirac measures, and it is not in $L^2$, meaning that $\langle x|x\rangle$ is not defined. In other words, states concentrated at a single point is not allowed in the theory from a strict mathematical point of view. So 4 is not true. If you want to include states like $|x\rangle$ there is a formalism called ...


0

There are two mathematically rigorous (and quite general) Hilbert Space formalisms you might be looking for. Both can be seen as attempts to salvage the engine of Dirac's original bra-ket algorithm, while avoiding its mathematical embarrassments. The first - created by von Neumann - replaces Dirac's kets by vectors in an abstract Hilbert space $\mathscr{H}$,...


1

Without orthogonality this is false: an example is given by Robert Israel. Orthogonality implies that $\|u+v\|^2 = \|u\|^2+\|v\|^2$ for $u\in M$, $v\in N$. Thus, if a sequence $(u_n+v_n)$ converges, the inequalities such as $$\|u_n-u_m\|\le \|(u_n+v_n)-(u_m+v_m)\|$$ imply convergence of both $u_n$ and $v_n$. So, $u_n\to u\in M$ and $v_n\to v\in N$, which ...


0

Recall Gelfand's formula for the spectral radius of a bounded operator $T$: $$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n}. $$ If $T$ is a self-adjoint operator and $\|f\|=1$ then $$\|Tf\|^2 = \langle Tf, Tf\rangle = \langle T^2f, f\rangle\leqslant\|T^2f\|\|f\|=\|T^2f\| $$ which implies $\|T^2\|=\|T\|^2$. By induction it follows that $\|T^{2^n}\|=\|T\|^{2^n}$ ...


0

Please see the following rather self-contained proof: Link: http://people.math.gatech.edu/~heil/6338/summer08/section5a_adjoint.pdf


4

Here is a hint: show that $K_u(R_0)$ is both closed and convex. Then apply a result your professor proved - or is in your book - about closed and convex sets to conclude that $K_u(R_0)$ contains a unique point that is closest to the origin in $H$.


1

(1). For a Banach space $B,$ with scalars $S=\mathbb R$ or $S=\mathbb C,$ the weak$^*$ topology on $B^*$ is defined to be the weakest topology on the set $B^*$ such that for every $x\in B$ the function $$\bar x:B^*\to S$$ is continuous, where $$\bar x (f)=f(x).$$ So it is necessary and sufficient that $\bar x^{-1}U=\{f\in B^*:f(x)\in U\}$ is a weak$^*$-open ...


2

I suggest the following link http://mathforum.org/library/drmath/view/74532.html. One can construct the hyperplane that bisects the two vectors, then the matrix $U$ is the reflection with respect to this hyperplane. For the given problem, $x-y$ is perpendicular to this hyperplane. One can show this by $(x-y)\cdot(x+y)=0$. Then for $v=(x-y)/||x-y||$, the ...


2

If you are just looking for existence, consider the span $S = \operatorname{span} \{x,y\}$, take two orthogonal vectors $u_1,u_2\in S$, and using Gram-Schmidt extend this set to an orthogonal basis $u_1,u_2,\ldots,u_n$ of $\mathbb R^n$. Now Consider the matrix $$U=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\\&&1\\&&&...


1

$\newcommand\ip[2]{\langle#1,#2\rangle}$ It's false. For example define $T:\Bbb C^2\to\Bbb C^2$ by $$T(x_1,x_2)=(x_2,0);$$then $||T||=1$ while the AM-GM inequality followed by Holder's inequality shows that $$|\ip {Tx}x|\le\frac{||x||^2}{2}.$$ It should probably be noted that it's less trivially false in the complex case than in the real case. As has been ...


0

Let $x,y \in X$, then we have \begin{align*} \sum_{k=0}^3 i^k B(x+i^k y, x+i^ky) &= B(x+y, x+y) - B(x-y,x-y) \\ & \quad + i \left[B(x+iy,x+iy) - B(x-iy, x-iy) \right] \\ &= 2 B(y,x)+2B(x,y) + 2B(x,y)-2B(x,y) + 2 B(y,x) \\ &= 4B(y,x). \end{align*} Since $ B(x+i^k y, x+i^ky) \in \mathbb{R}$, we have \begin{align*} 4\overline{B(y,x)} &=\...


0

Let $\lambda\notin \sigma(A)$ then $A-\lambda 1$ is invertible, denote it's inverse by $C\in B(H_2)$. We have $C(A-\lambda1)=(A-\lambda1)C=1$. Therefore, $D=U^{-1}CU\in B(H_1)$ is the inverse of $(B-\lambda1)$ (why? write $B=U^{-1}AU$). Thus, $\lambda \notin \sigma(B)$. Now, if $\lambda\in \sigma_p(A)$ that means $\exists v\ne 0$ such that $Av=\lambda v$. ...


1

You are almost done. Since $f_n\to f$ pointwise, and $f_n'\to h$ in $L^2$, it follows that for every $a,b\in\mathbb{R}$, $$f(b)-f(a)= \lim_{n\to\infty}(f_n(b)-f_n(a)) = \lim_{n\to\infty}\int_a^b f_n'(t)\,dt = \int_a^b h(t)\,dt $$ The last step here is based on the Cauchy-Schwarz inequality, $$\int_a^b |f_n'(t)-h(t)|\,dt \le (b-a)^{1/2}\left(\int_a^b |f_n'(t)-...


2

For every $x$ we have $x=Px+(x-Px)$ where $Px\in \operatorname{im}P$ and $(x-Px)\in \ker P$. So, $\ker P$ is a complement of $\operatorname{im} P$. Suppose it is not an orthogonal complement; then there exist $u\in \operatorname{im}P$ and $v\in \ker P$ such that $\operatorname{Re}\langle u, v \rangle < 0$. For sufficiently small $t>0$ we have $\|u+tv\...


1

The GNS construction doesn't have a canonical way of choosing "the right amount" of states. For an arbitrary algebra one uses all states (see for instance Corollary I.9.11 in Davidson's C$^*$-Algebras by Example). But this is often way too much. Let's think of the easiest concrete example: let $A=M_2(\mathbb C)$. If we follow the abstract recipe, we need ...


1

Without loss of generality, assume that $\lambda_k \ne 0$ for all $k$. Otherwise, discard $v_k$ from your orthonormal set, and discard the term $\lambda_k (x,v_k)v_k$ from the defining sum for $F$. Nothing is changed by making such changes. And assume that $\{ v_k \}$ is an orthonormal set by discarding $0$ vectors and renormalizing. The largest domain $\...


2

Assuming $\{ e_n \}$ and $\{ e_n'\}$ are orthonormal, define $$ Lx = \sum_{n=1}^{\infty}(x,e_n)e_n' $$ The operator $L$ is linear and satisfies $$ \|Lx\|^2=\sum_{n=1}^{\infty}|(x,e_n)|^2 \le \|x\|^2. $$ Now assume that $\{e_n\}$ is a complete orthonormal basis, and assume that $d=\left(\sum_{n=1}^{\infty}\|e_n-e_n'\|^2\right)^{1/2} ...


1

Pedestrian way.. Eigenvector: $${\sum}_k(\lambda-\lambda_k)\alpha_kv_k=\lambda({\sum}_k\alpha_kv_k)-{\sum}_k\alpha_k(\lambda_kv_k)=\\=F({\sum}_k\alpha_kv_k)-{\sum}_k\alpha_k(Fv_k) =F({\sum}_k\alpha_kv_k-{\sum}_k\alpha_kv_k)=F(0)=0$$ Nullvector: $${\sum}_k|\lambda-\lambda_k|^2|\alpha_k|^2=\|{\sum}_k(\lambda-\lambda_k)\alpha_kv_k\|^2=\|0\|=0$$ The rest ...


1

This can be done by a straightforward calculation: For all $x \in X$ and $y \in Y$ we have \begin{align*} \langle x, S(y) \rangle &= \langle x, J_X^{-1}(T'(J_Y(y))) \rangle = \langle x, J_X^{-1}(T'( \langle y, - \rangle)) \rangle = \langle x, J_X^{-1}(\langle y, T(-) \rangle) \rangle \\ &= \overline{\langle J_X^{-1}(\langle y, T(-) \...


1

This operator is the orthogonal projection onto $U$, it holds $C^2=C$: First we find that $$ \langle Cv,e_n\rangle = \langle v,e_n\rangle, $$ since $(e_n)$ is orthonormal. This implies $$ C^2v = \sum_n \langle Cv,e_n\rangle Ce_n = \sum_n \langle v,e_n\rangle e_n =Cv. $$ Hence, $C^2 = C$, which implies $C^{1/2}=C$. Edit: I assumed that the scalar product ...


0

You should know that for each $f \in L^{p}_{loc}(\Omega)$ we can define a Distribution by: $T_{f}: \mathcal{D}(\Omega) \rightarrow\mathbb{R}$, where $T_{f}(\phi) = \displaystyle\int_{\Omega}f\phi$. In the case, $T_{f}$ is called regular distribution. The Du Bouis Reymond Lemma states that the map $f\mapsto T_{f}$ is one-to-one. However, there are ...


1

The fact that $\iota$ is HS means that if $\{e_n\}$ is an orthonormal basis of $U$, then $$\tag{1} \sum_n\langle e_n,e_n\rangle_V=\sum_n\|\iota(e_n)\|_V^2<\infty. $$ After identifying $U$ and $V$ with their respective duals, we have $C:V\to U$ given by $$\tag{2} \langle Cv,u\rangle_U=\langle v,\iota (u)\rangle_V=\langle v,u\rangle_V. $$ In particular, ...


2

You can write unravel this definition by writing \begin{align} Af(x) & = \int_{0}^{x}K(x,y)f(y)dy+\int_{x}^{1}K(x,y)f(y)dy \\ & =\sinh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\sinh(x)\int_{x}^{1}\sinh(1-y)f(y)dy. \end{align} This is a typical kind of Green function expression. First note that $$ (Af)(0) = 0,\;\;\; (Af)(1)=0. $$ Then, \begin{...


2

Weak convergence means: for every $f$ the sequence $\{f(x_n)\}$ tends to $0$. When you are proving a statement that begins with "for every $f$", you don't get to choose $f$. So, there is no reason for $\|f\|$ to be small. Your argument doesn't work. Instead, use Bessel's inequality $\sum |\langle y_f, x_n\rangle|^2 \le \|y_f\|^2$, and the fact that the ...


1

Starting with the half of the question that no longer exists: $M$ invertible does not imply $T$ invertible. Say $\phi=e_{-2}$. Then $M$ is invertible. But if $f=e_1$ then $Tf=P(e_{-1})=0$, so $T$ is not invertible. On the other hand $T$ invertible does imply $M$ invertible. Suppose $M$ is not invertible. Let $\epsilon>0$, and set $$E=\{x:|\phi(x)|<\...


2

$W_C = \{u + iv \in X_C \mid u, v \in W\}$ is invariant under $T_C$, so we may as well forget about the real Banach spaces and consider an operator $T$ on a complex Banach space $X$ with a finite-codimensional invariant subspace $W$, such that $\sigma(T) \subseteq \mathbb R$. The claim is that $\sigma(T|_W) \subseteq \mathbb R$. If $\lambda \notin \sigma(T)...


0

No, that's not a good definition of an orthogonal basis. It's not enough to say every vector can be written as a linear combination, because you may also need infinite sums, which are no longer linear combinations. The sentence "other vectors not in the Hilbert space may also be written as a linear combination of these vectors." makes no sense. What are ...


2

Looking at the definition of a norm on wikipedia (which matches the definitions I encountered in textbooks) (link here) if $V$ is a vector space, then a norm is a function $\rho: V\to \mathbb R$ that satisfies certain properties so it is quite explicitly stated that the norm of every vector is a real number, thus finite.


1

This is exactly the uniqueness in the polar decomposition. You have, since $v $ is a partial isometry, $$\tag {2}{\text {ran}\,v^*v}= {\text {ran}\,v^*}=(\ker v)^\perp=(\ker y)^\perp=\overline {\text {ran}\,y}. $$ Suppose that $w,z $ gives another such decomposition of $x $. Let $p=v^*v=w^*w $. Then, since $py=y$, we have $$ y^2=y^*y=y^*py=y^*v^*vy=x^*x=|x|...


1

Any bounded sequence in a separable Hilbert space (which is reflexive) has a weakly convergent subsequence. Added on edit: See Theorem 3.18 of the same book.


0

Hint: use the Riesz theorem: Let $y\in R$, the linear function defined on $H$ by $f_y(x)=<T(x),y>$ is continuous, thus you can find $T^*(y)\in H$ such that $f_y(x)=<x,T^*(y)>$.


-1

This is (tightly related to) formulae (6.11.2) in Maz'ya's book on Sobolev spaces, 2011 edition; or Corollary 1 in §4.11.1 in the 1985 edition.


1

Say $x \in V$. Since the $e_n$ are a basis for $H$ and $x \in H$, $x = \sum_n c_n e_n$ for some coefficients $c_n \in \mathbb{K}$. Since $Q$ and $1-Q$ are orthogonal projections whose sum is 1, $x = Qx + (1-Q)x = (Q \sum_n c_n e_n) + ((1-Q) \sum_n c_n e_n)$. But the second term is zero since $x \in V$ implies $Qx = x$. So we have $x = Q \sum_n c_n e_n$, and ...


1

$$T^2x=\sum_n\lambda_n\langle Tx,\phi_n\rangle\phi_n=\sum_n\lambda_n\left\langle \sum_m \lambda_m\langle x,\phi_m\rangle\phi_m,\phi_n\right\rangle \phi_n\\ =\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle\langle\phi_m,\phi_n\rangle \phi_n=\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle||\phi_n||^2\delta_{m,n} \phi_n\\ =\sum_n\lambda_n\lambda_n\...


2

Usually, $(x_k)_{k \in \Bbb N}$ is called an orthogonal sequence if $\langle x_j, x_k \rangle = 0$ whenever $j \neq k$.


1

Continuous: Note that $$ \|Au\|^2 = \sum_k \left(\sum_h a_{k,h} u_h\right)^2 \leq \sum_k \left(\left(\sum_h a_{k,h}^2\right) \|u\|^2\right) $$ Not Compact: Note that $(Ae_n)_k = a_{k,n}$. Thus, we note that for any $n$, we have $(Ae_n)_{n-1} = 1$, so that $\|Ae_n\| \geq 1$. Conclude that $A e_n \not \to 0$.


0

$(3)$ must not hold for $u=y-v$ (only if $y\in U$). That was my mistake. Let $U$ and $H$ be Hilbert spaces and $\iota$ be an embedding of $U$ into $H$. Then, $$\pi x:=u\;\;\;\text{for }x\in H\text{ with }x=\iota u+y\text{ for some }u\in U\text{ and }y\in\left(\iota U\right)^\perp$$ is a well-defined mapping $H\to U$. If $\iota$ is an isometry, then $$\...



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